measure countably additive and closure
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In Probability and Measure Theory by Ash I, pag. 23, I found next propositions:
$F(R)$: field of finite disjoint unions of right-semiclosed intervals of R.
F is distribution function.
Let A1, A2, A3... be a sequence of sets in $F(R)$ decreasing to empty set. If (a,b] is one of the intervals of An, then by right continuity of F,
mu(a',b] = F(b)-F(a')-->mu(a,b] = F(b)-F(a) as a-->a' from above.
after that, Ash says "we can find set Bn in F(R) whose closures(Bn) are included in An, with mu(Bn) approximating mu(An)"
I think, Bn is a rigth-semiclosed interval, like (a,b] so, closure((a,b]) = [a,b] and this last is not included in An because An is in form of (a,b]
I think it is a mistake.:
here is a imagen of this
probability
asked Nov 15 at 16:07
Aldo RM
61
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Aldo RM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
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In Probability and Measure Theory by Ash I, pag. 23, I found next propositions:
$F(R)$: field of finite disjoint unions of right-semiclosed intervals of R.
F is distribution function.
Let A1, A2, A3... be a sequence of sets in $F(R)$ decreasing to empty set. If (a,b] is one of the intervals of An, then by right continuity of F,
mu(a',b] = F(b)-F(a')-->mu(a,b] = F(b)-F(a) as a-->a' from above.
after that, Ash says "we can find set Bn in F(R) whose closures(Bn) are included in An, with mu(Bn) approximating mu(An)"
I think, Bn is a rigth-semiclosed interval, like (a,b] so, closure((a,b]) = [a,b] and this last is not included in An because An is in form of (a,b]
I think it is a mistake.:
here is a imagen of this
probability
asked Nov 15 at 16:07
Aldo RM
61
New contributor
Aldo RM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In Probability and Measure Theory by Ash I, pag. 23, I found next propositions:
$F(R)$: field of finite disjoint unions of right-semiclosed intervals of R.
F is distribution function.
Let A1, A2, A3... be a sequence of sets in $F(R)$ decreasing to empty set. If (a,b] is one of the intervals of An, then by right continuity of F,
mu(a',b] = F(b)-F(a')-->mu(a,b] = F(b)-F(a) as a-->a' from above.
after that, Ash says "we can find set Bn in F(R) whose closures(Bn) are included in An, with mu(Bn) approximating mu(An)"
I think, Bn is a rigth-semiclosed interval, like (a,b] so, closure((a,b]) = [a,b] and this last is not included in An because An is in form of (a,b]
I think it is a mistake.:
here is a imagen of this
probability
asked Nov 15 at 16:07
Aldo RM
61
New contributor
Aldo RM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In Probability and Measure Theory by Ash I, pag. 23, I found next propositions:
$F(R)$: field of finite disjoint unions of right-semiclosed intervals of R.
F is distribution function.
Let A1, A2, A3... be a sequence of sets in $F(R)$ decreasing to empty set. If (a,b] is one of the intervals of An, then by right continuity of F,
mu(a',b] = F(b)-F(a')-->mu(a,b] = F(b)-F(a) as a-->a' from above.
after that, Ash says "we can find set Bn in F(R) whose closures(Bn) are included in An, with mu(Bn) approximating mu(An)"
I think, Bn is a rigth-semiclosed interval, like (a,b] so, closure((a,b]) = [a,b] and this last is not included in An because An is in form of (a,b]
I think it is a mistake.:
here is a imagen of this
probability
probability
asked Nov 15 at 16:07
Aldo RM
61
New contributor
Aldo RM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 16:07
Aldo RM
61
New contributor
Aldo RM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 16:07
Aldo RM
61
New contributor
Aldo RM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 16:07
Aldo RM
61
asked Nov 15 at 16:07
Aldo RM
61
61
New contributor
Aldo RM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Aldo RM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Aldo RM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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Aldo RM is a new contributor. Be nice, and check out our Code of Conduct.
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